Consider the following elementary reaction: CFCl3(g)→CFCl2(g)+Cl(g).Suppose we let k1 stand for the rate constant of this reaction, and k−1 stand for the rate constant of the reverse reaction.

Write an expression that gives the equilibrium concentration of CFCl3 in terms of k1,k−1, and the equilibrium concentrations of CFCl2 and Cl.

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Edufirst Edufirst · Answer 1 · 2020-04-14T00:33:34+02:00

Answer:

[tex][CFCl_3(g)]=\dfrac{k_{-1}}{k_1}\cdot[CFCl_2(g)]\cdot [Cl(g)][/tex]

Explanation:

The equilibrium constant is equal to the ratio of the rate constant of the forward reaction to the rate constant of the reverse reaction:

[tex]K_c=\dfrac{k_{forward}}{k_{reverse}}[/tex]

Then, using k₁ and k₋₁ for the rate constants of the forward and the reverse reactions, respectively:

[tex]k_c=\dfrac{k_1}{k_{-1}}[/tex]

The equilibrium equation is:

CFCl₃(g) ⇄ CFCl₂(g) + Cl(g)

For which the equilibrium constant is:

[tex]k_c=\dfrac{[CFCl_2(g)]\cdot [Cl(g)]}{[CFCl_3(g)]}=\dfrac{k_1}{k_{-1}}[/tex]

Now you can write the equilibrium concentraion of CFCl₃(g) in terms of k₁, k₋₁, [CFCl₂(g)], and [Cl(g)]:

[tex][CFCl_3(g)]=\dfrac{k_{-1}}{k_1}\cdot[CFCl_2(g)]\cdot [Cl(g)][/tex]